Prakhar Gupta has 4+ years of experience in Content Marketing Field. 

Content Marketing Specialist

Number System questions are a very common type of questions asked in almost every competitive exam. These questions carry a weightage of 2-3 questions (4-6 marks) in SSC exams. To get a good rank in competitive exams, you should have a great practice of variety of questions of Number system

Here are some tips for solving Number System questions: Understand prime numbers, factors, multiples, LCM, and HCF thoroughly. Practice divisibility rules for quick checks. Tackle questions systematically, eliminating wrong choices in multiple-choice questions. Practice with previous exam papers to familiarize yourself with question patterns.

So, we have attached 10 questions of Number System for you to practice with. You should aim to solve these questions in less than half a minute for each.

 

<h2 id="content-id-1702221920303-0">Practice Questions on Number System</h2>

You can also download the Number System questions and answers pdf. Just click on the Download PDF button. So let's start with the very first question.

Q:1 Find the number of factors of the numbers 225 and 525

1. 8 ; 9

2. 9 ; 8

3. 9 ; 9

4. 8 ; 8

(Difficulty: 3, Estimated Time: 20 Seconds) A basic one! These type of questions are very common

Q:2 There are three different sizes 75 cm, 225 cm, and 125 cm of the thread of equal size to cut from each thread. Find the greatest possible value such that no part thread is left.

1. 30 cm

2. 21 cm

3. 25 cm

4. 35 cm

(Difficulty: 3, Estimated Time: 20 Seconds) Try using short tricks, they will save your time

Q:3 The ratio of any two numbers is 6 : 7 and the LCM of the numbers is 168. Find the difference between them.

1. 4

2. 28

3. 16

4. 8

(Difficulty: 2, Estimated Time: 10 Seconds) This was an easy one! Did you get it right?

Q:4 Which of the following fraction does not lie between 3/8 and 7/12?

1. 5/12

2. 3/5

3. 5/9

4. 4/11

(Difficulty: 3, Estimated Time: 20 Seconds) This was a test of your concepts!

Q:5 What is the H.C.F. of 1057, 1661, 4681?

1. 153

2. 161

3. 151

4. None of these

(Difficulty: 3, Estimated Time: 20 Seconds) We're halfway through. Have you got all your questions correct so far?

Q:6 Find the number which is completely divisible by 7 from the following:

403487, 87247 and 2247861

1. 403487 and 87247

2. 87247 and 2247861

3. 403487, 87247 and 2247861

4. 403487 and 2247861

(Difficulty: 4, Estimated Time: 30 Seconds) This was a hard nut to crack, be prepared for such questions in exam!

Q:7 Find the product of LCM and HCF of given numbers as follow:

12/27, 10/9, 15/24, 36/5, and 22/9.

1. 1/175

2. 11/6

3. 6/11

4. 175

(Difficulty: 3, Estimated Time: 20 Seconds) If you have a good understanding, you might have wrapped it up in 10 seconds!

Q:8 Find the least number which when added to 115 makes it divisible by 3, 5, 6 and 8.

1. 5

2. 6

3. 3

4. 0

(Difficulty: 2, Estimated Time: 15 Seconds) This was an interesting question!

Q:9 If (b = c), then find the LCM of given numbers:

a2bc, b3ca, a8b5, c9ab2, cab5 and c2a3b

1. a5b9

2. a8b9

3. a8b9c

4. a8b11 

(Difficulty: 3, Estimated Time: 20 Seconds) Be ready for such questions in your exams!

Q:10 If 9834X11 is divisible by 11 then, what will be the possible value of X if X is a single digit?

1. 4

2. 2

3. 0

4. 3

(Difficulty: 2, Estimated Time: 15 Seconds) This was an easy one. Did you guess them all correctly?

<h2 id="content-id-1702221920330-1">Answer Key</h2>

Let's check out your score in this test.

<table border="0" cellpadding="1" cellspacing="1">
	<tbody>
		<tr>
			<td>1. (3)</td>
			<td>2. (3)</td>
			<td>3. (1)</td>
			<td>4. (2)</td>
			<td>5. (3)</td>
		</tr>
		<tr>
			<td>6. (4)</td>
			<td>7. (2)</td>
			<td>8. (1)</td>
			<td>9. (4)</td>
			<td>10. (3)</td>
		</tr>
	</tbody>
</table>

Comment below your score, considering each question has 1 mark only. If you scored 8 to 10, congratulations! You are one step closer to selection. If you have scored 5 to 8 marks, then you are doing well, keep it up. If you have scored less than 5 marks then you need to work a little harder on this subject. But don't worry, we are here to help you master the subject.

Let's check the answers and solutions and try to find out what went wrong. 
 

<h2 id="content-id-1702221920330-2">Answers and Solutions</h2>

Q:1 The correct answer is Option 3 i.e. 9 ; 9.

A factor divides a given number completely without leaving any remainder.

225 is divisible by 1, 3, 5, 9, 15, 25, 45, 75, 225.

525 is divisible by 1, 3, 5, 7, 15, 21, 35, 105, 525.

Hence, number of factors of 225 and 525 is 9 each.

 

Q:2 The correct answer is option 3 i.e. 25 cm.

The highest possible value is obtained after taking the H.C.F of given lengths

H.C.F of 75, 225, 125 is equal to 25

So, The greatest possible value is 25 cm.

 

Q:3 The correct answer is option 1 i.e. 4

Let the numbers be 6x and 7x 

Then its LCM will be 42x 

A/Q, 42x = 168

x = 4

So, the numbers are 24 and 28

The difference between numbers = 28 - 24 = 4

 

Q:4 The correct answer is Option 2 i.e. 3/5.

By converting the fractions into percentages,

⇒ 3/8 = 3(1/8) = 3 × 12.5% = 37.5%

⇒ 7/12 = 7(1/12) = 7 × 8.33% = 58.33%

Now, we have to check which of the following fractions does not lie between the given values

⇒ 5/12 = 5(1/12) = 5 × 8.33% = 41.67%

⇒ 3/5 = 3(1/5) = 3 × 20% = 60%

 

Q:5 The correct answer is Option 3 i.e. 151.

⇒ 1057 = (7 × 151)

⇒ 1661 = (11 × 151)

⇒ 4681 = (31 × 151)

The highest common factor of these numbers is 151

 

Q:6 The correct answer is Option 4 i.e. 403487 and 2247861.

Divisibility rule of 7: Take the unit digit of the number, multiply it by 2, and subtract from the rest of the number. Repeat till 2-digit number

403487, 87247 and 2247861

(i) 403487: 40348 - (7 × 2) = 40334

⇒ 40334: 4033 - (4 × 2) = 4025

⇒ 4025: 402 - (5 × 2) = 392

⇒ 392: 39 - (2 × 2) = 35 

Here, 35 is completely divisible by 7 so, 403487 also divisible by 7

(ii) 87247: 8724 - (7 × 2) = 8710

⇒ 8710: 871 - (0 × 2) = 871

⇒ 871: 87 - (1 × 2) = 85

Here, 85 is not divisible by 7 so, 87247 also not divisible by 7

(iii) 2247861: 224786 - (1 × 2) = 224784

⇒ 224784: 22478 - (4 × 2) = 22470

⇒ 22470: 2247 - (0 × 2) = 2247

⇒ 2247: 224 - (7 × 2) = 210

⇒ 210: 21 - (0 × 2) = 21

Here, 21 is completely divisible by 7 so, 2247861 also divisible by 7

Here,

403487 and 2247861 are completely divisible by 7

 

Q:7 The correct answer is option 2 i.e. 11/6

12/27, 10/9, 15/24, 36/5, and 22/9

Numerator numbers = 12, 10, 15, 36, 22

12 = 22 × 3

10 = 2 × 5

15 = 3× 5

36 = 22 × 32

22 = 2 × 11

Now, Denominator numbers = 27, 9, 24, 5, 9

27 = 33

9 = 32

24 = 23 × 3

5 = 5

9 = 32

LCM of Numerators = 22 × 32 × 5 × 11 = 4 × 9 × 5 × 11 = 1980

HCF of Numerators = 1

LCM of Denominators = 23 × 33 × 5 = 8 × 27 × 5 = 1080

HCF of Denominators = 1

Formula:

LCM of factions = [(LCM of Numerators)/(HCF of Denominators)]

And,

HCF of factions = [(HCF of Numerators)/(LCM of Denominators)]

LCM of 12/27, 10/9, 15/24, 36/5, and 22/9 = 1980/1 = 1980

And,

HCF of 12/27, 10/9, 15/24, 36/5, and 22/9 = 1/1080

The product of LCM and HCF is,

(1980) × (1/1080) = 198/108 = 99/54 = 11/6

 

Q:8 The correct answer is option 1 i.e. 5.

The least number must be the LCM of given numbers

The LCM of 3, 5, 6, and 8 = 120

So, the least number which is added to 115 is 5 so it is divided by 3, 5, 6, and 8

The possible least value is 5.

 

Q:9 The correct answer is Option 4 i.e. a8b11. 

Numbers = a2bc, b3ca, a8b5, c9ab2, cab5 and c2a3b

(b = c)

So,

⇒ a2bc = a2bb = a2b2

⇒ b3ca = b3ba = ab4

⇒ a8b5 = a8b5

⇒ c9ab2 = b9ab2 = ab11

⇒ cab5 = bab5 = ab6 

⇒ c2a3b = b2a3b = a3b3

LCM of a2b2, ab4, a8b5, ab11, ab6, a3b5 = a8b11 

 

Q:10 The correct answer is Option 3 i.e. 0.

A natural number is divisible by 11 if the difference between the sum of the odd-numbered digits and the sum of even-numbered digits is a multiple of 11 or 0

The given number is 9834X11

The digit at odd places 9, 3, X, 1

The sum of these digits = 9 + 3 + X + 1 = 13 + X

The digits at even place are 8, 4, 1 

The sum of these digits = 13

The difference between the sums = 13 + X - 13 = X

So, the possible value of X = 0, 11, 22, 33................

But X is a single-digit

So, the X = 0

So, this is it for today. We will meet again with another new topic. Till then, you can practice the questions again by downloading the PDF of Number System.

Since 1995, Neetu Mam is Providing Free Content for English to the Students

Best Publisher for English Content

Practice Questions on Number System!

Number System Questions along with detailed Solutions PDF

No title

Latest Notifications

Exam Eligibility

Exam Dates

Exam Pattern

Syllabus

Admit Card

Exam Cutoff

Answer Key

Exam Results

Daily Current Affairs

Current Affairs Quiz

Daily Editorial

Daily Vocabulary

Study Plan

Important Days

Question Paper

Previous Year PDF

English Language

Logical Reasoning

Quantitative Aptitude

General Knowledge

General Awareness

General Science

Today in History

Know Your State

Full Form

Know Your Country

Government Schemes

Know Your City

Know Your Leader

Books and Authors

question answer

Quiz

Courses

Classes

Quant Tricks for SSC CGL Exam

&lt;p&gt;No title&lt;/p&gt;

Quant Important Topics for SSC CGL Exam

Check out the comprehensive set of Number System questions and answers, with detailed solutions in PDF. It is the perfect resource for anyone who wants to improve their problem solving skills for competitive exams.

Number System Questions along with detailed Solutions PDF

Number System FAQs

Number System Comments:

Number System RELATED ARTICLES